A $10$ -meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at $3$ meters per minute. At a certain instant, the bottom of the ladder is $6$ meters from the wall. What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $12$ (Choice B) B $-6$ (Choice C) C $-\dfrac72$ (Choice D) D $7$
Setting up the math Let... $a(t)$ denote the distance between the top of the ladder and the ground at time $t$, $b(t)$ denote the distance between the bottom of the ladder and the wall at time $t$, $c$ denote the length of the ladder (which is always $10$ meters), and $A(t)$ denote the area formed by the ladder at time $t$. $a(t)$ $b(t)$ $c$ $A(t)$ We are given that $c=10$ and $a'(t)=-3$. Notice that $a'$ is negative. We are also given that $b(t_0)=6$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures The sides relate to each other through the Pythagorean theorem: $\begin{aligned} \,[a(t)]^2+[b(t)]^2&=c^2 \\\\\\ [a(t)]^2+[b(t)]^2&=10^2 \end{aligned}$ We can differentiate both sides to find an expression for $b'(t)$ : $b'(t)=-\dfrac{a(t)a'(t)}{b(t)}$ The sides and the area relate to each other through the formula for right triangle area: $A(t)=\dfrac{a(t)b(t)}{2}$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=\dfrac{a'(t)b(t)+a(t)b'(t)}{2}$ Using the information to solve In order to find $b'(t_0)$ we need to find $a(t_0)$. Using the Pythagorean theorem and the fact that $b(t_0)=6$ and $c=10$, we can find that $a(t_0)=8$. Let's plug ${a(t_0)}={8}$, ${a'(t_0)}={-3}$, and ${b(t_0)}={6}$ into the expression for $b'(t_0)$ : $\begin{aligned} b'(t_0)&=-\dfrac{{a(t_0)}{a'(t_0)}}{{b(t_0)}} \\\\ &=-\dfrac{({8})({-3})}{({6})} \\\\ &=4 \end{aligned}$ Now let's plug ${a(t_0)}={8}$, ${a'(t_0)}={-3}$, ${b(t_0)}={6}$, and $C{b'(t_0)}=C{4}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=\dfrac{{a'(t_0)}{b(t_0)}+{a(t_0)}C{b'(t_0)}}{2} \\\\ &=\dfrac{({-3})({6})+({8})\left(C{4}\right)}{2} \\\\ &=7 \end{aligned}$ In conclusion, the rate of change of the area formed by the ladder at that instant is $7$ square meters per minute. Since the rate of change is positive, we know that the area is increasing.